Inverse/Reverse of Number of Permutations and of Number of Combinations with Repetitions? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to reverse the $n$ choose $k$ formula?Permutations with Repetitions, how to select $n$ and $r$How many possible combinations/permutations?Trouble with permutations and combinations problemsCombinatorics: terminology for permutations and combinationsTerminology clarification: are these permutations or combinations?Finding $n$ permutations $r$ with repetitionsCombinations and Permutations artistHow can we count combinations with repetition (or permutations) using inequality symbols between each number?Visualizing combinations with repetitions allowed.Using the numbers 1, 2, 3, 4 and 5, how many 3-digit combinations can you make? Repetitions allowed.
What does できなさすぎる means?
How come Sam didn't become Lord of Horn Hill?
How does the math work when buying airline miles?
What would be the ideal power source for a cybernetic eye?
How to answer "Have you ever been terminated?"
Trademark violation for app?
If a VARCHAR(MAX) column is included in an index, is the entire value always stored in the index page(s)?
Is it a good idea to use CNN to classify 1D signal?
Why are there no cargo aircraft with "flying wing" design?
What is homebrew?
Is "Reachable Object" really an NP-complete problem?
If a contract sometimes uses the wrong name, is it still valid?
Why are the trig functions versine, haversine, exsecant, etc, rarely used in modern mathematics?
How can I use the Python library networkx from Mathematica?
Most bit efficient text communication method?
Delete nth line from bottom
An adverb for when you're not exaggerating
Fantasy story; one type of magic grows in power with use, but the more powerful they are, they more they are drawn to travel to their source
Why aren't air breathing engines used as small first stages
Can you use the Shield Master feat to shove someone before you make an attack by using a Readied action?
Compare a given version number in the form major.minor.build.patch and see if one is less than the other
Chinese Seal on silk painting - what does it mean?
How to convince students of the implication truth values?
Can a new player join a group only when a new campaign starts?
Inverse/Reverse of Number of Permutations and of Number of Combinations with Repetitions?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to reverse the $n$ choose $k$ formula?Permutations with Repetitions, how to select $n$ and $r$How many possible combinations/permutations?Trouble with permutations and combinations problemsCombinatorics: terminology for permutations and combinationsTerminology clarification: are these permutations or combinations?Finding $n$ permutations $r$ with repetitionsCombinations and Permutations artistHow can we count combinations with repetition (or permutations) using inequality symbols between each number?Visualizing combinations with repetitions allowed.Using the numbers 1, 2, 3, 4 and 5, how many 3-digit combinations can you make? Repetitions allowed.
$begingroup$
For an engineering application, I need the inverse functions of the computations of the number of combinations and permutations.
In the thread How to reverse the $n$ choose $k$ formula? it shows how to reverse/inverse the number of combinations function...
Given:
$X = C(n, k) = binomnk = fracn! k!(n-k)! $
then you can limit n to:
$sqrt[k]k! X + k ge n ge sqrt[k]k! X$
And thus you at most need to check k+1 possible values of n.
Great!
But how do I solve for k instead of for n?
And given instead:
$X = P(n, k) = fracn! (n-k)! $
then would I be correct in assuming from above that:
$sqrt[k]X + k ge n ge sqrt[k]X$
? And if so, how would I then solve for k instead of n?
And finally, given instead CR is Combinations with Repetitions:
$X = CR(n, k) = binomn+k-1k = binomn+k-1n-1 = frac (n+k-1)! k!(n-1)! $
then would I be correct in assuming from above that:
$sqrt[k]k! X + 1 ge n ge sqrt[k]k! X - k + 1$
? And if so, how would I then solve for k instead of n?
Any help on the algebra for these five related inverse functions would be greatly appreciated!
combinatorics permutations combinations binomial-coefficients inverse-function
$endgroup$
add a comment |
$begingroup$
For an engineering application, I need the inverse functions of the computations of the number of combinations and permutations.
In the thread How to reverse the $n$ choose $k$ formula? it shows how to reverse/inverse the number of combinations function...
Given:
$X = C(n, k) = binomnk = fracn! k!(n-k)! $
then you can limit n to:
$sqrt[k]k! X + k ge n ge sqrt[k]k! X$
And thus you at most need to check k+1 possible values of n.
Great!
But how do I solve for k instead of for n?
And given instead:
$X = P(n, k) = fracn! (n-k)! $
then would I be correct in assuming from above that:
$sqrt[k]X + k ge n ge sqrt[k]X$
? And if so, how would I then solve for k instead of n?
And finally, given instead CR is Combinations with Repetitions:
$X = CR(n, k) = binomn+k-1k = binomn+k-1n-1 = frac (n+k-1)! k!(n-1)! $
then would I be correct in assuming from above that:
$sqrt[k]k! X + 1 ge n ge sqrt[k]k! X - k + 1$
? And if so, how would I then solve for k instead of n?
Any help on the algebra for these five related inverse functions would be greatly appreciated!
combinatorics permutations combinations binomial-coefficients inverse-function
$endgroup$
1
$begingroup$
Question: when we are solving for $k$, is $n$ fixed?
$endgroup$
– Eevee Trainer
Mar 27 at 7:52
$begingroup$
Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
$endgroup$
– Brian Kennedy
Mar 27 at 15:15
$begingroup$
I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
$endgroup$
– Mike Earnest
Mar 28 at 16:29
$begingroup$
Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
$endgroup$
– Mike Earnest
Mar 28 at 16:33
add a comment |
$begingroup$
For an engineering application, I need the inverse functions of the computations of the number of combinations and permutations.
In the thread How to reverse the $n$ choose $k$ formula? it shows how to reverse/inverse the number of combinations function...
Given:
$X = C(n, k) = binomnk = fracn! k!(n-k)! $
then you can limit n to:
$sqrt[k]k! X + k ge n ge sqrt[k]k! X$
And thus you at most need to check k+1 possible values of n.
Great!
But how do I solve for k instead of for n?
And given instead:
$X = P(n, k) = fracn! (n-k)! $
then would I be correct in assuming from above that:
$sqrt[k]X + k ge n ge sqrt[k]X$
? And if so, how would I then solve for k instead of n?
And finally, given instead CR is Combinations with Repetitions:
$X = CR(n, k) = binomn+k-1k = binomn+k-1n-1 = frac (n+k-1)! k!(n-1)! $
then would I be correct in assuming from above that:
$sqrt[k]k! X + 1 ge n ge sqrt[k]k! X - k + 1$
? And if so, how would I then solve for k instead of n?
Any help on the algebra for these five related inverse functions would be greatly appreciated!
combinatorics permutations combinations binomial-coefficients inverse-function
$endgroup$
For an engineering application, I need the inverse functions of the computations of the number of combinations and permutations.
In the thread How to reverse the $n$ choose $k$ formula? it shows how to reverse/inverse the number of combinations function...
Given:
$X = C(n, k) = binomnk = fracn! k!(n-k)! $
then you can limit n to:
$sqrt[k]k! X + k ge n ge sqrt[k]k! X$
And thus you at most need to check k+1 possible values of n.
Great!
But how do I solve for k instead of for n?
And given instead:
$X = P(n, k) = fracn! (n-k)! $
then would I be correct in assuming from above that:
$sqrt[k]X + k ge n ge sqrt[k]X$
? And if so, how would I then solve for k instead of n?
And finally, given instead CR is Combinations with Repetitions:
$X = CR(n, k) = binomn+k-1k = binomn+k-1n-1 = frac (n+k-1)! k!(n-1)! $
then would I be correct in assuming from above that:
$sqrt[k]k! X + 1 ge n ge sqrt[k]k! X - k + 1$
? And if so, how would I then solve for k instead of n?
Any help on the algebra for these five related inverse functions would be greatly appreciated!
combinatorics permutations combinations binomial-coefficients inverse-function
combinatorics permutations combinations binomial-coefficients inverse-function
asked Mar 27 at 6:52
Brian KennedyBrian Kennedy
1062
1062
1
$begingroup$
Question: when we are solving for $k$, is $n$ fixed?
$endgroup$
– Eevee Trainer
Mar 27 at 7:52
$begingroup$
Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
$endgroup$
– Brian Kennedy
Mar 27 at 15:15
$begingroup$
I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
$endgroup$
– Mike Earnest
Mar 28 at 16:29
$begingroup$
Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
$endgroup$
– Mike Earnest
Mar 28 at 16:33
add a comment |
1
$begingroup$
Question: when we are solving for $k$, is $n$ fixed?
$endgroup$
– Eevee Trainer
Mar 27 at 7:52
$begingroup$
Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
$endgroup$
– Brian Kennedy
Mar 27 at 15:15
$begingroup$
I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
$endgroup$
– Mike Earnest
Mar 28 at 16:29
$begingroup$
Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
$endgroup$
– Mike Earnest
Mar 28 at 16:33
1
1
$begingroup$
Question: when we are solving for $k$, is $n$ fixed?
$endgroup$
– Eevee Trainer
Mar 27 at 7:52
$begingroup$
Question: when we are solving for $k$, is $n$ fixed?
$endgroup$
– Eevee Trainer
Mar 27 at 7:52
$begingroup$
Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
$endgroup$
– Brian Kennedy
Mar 27 at 15:15
$begingroup$
Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
$endgroup$
– Brian Kennedy
Mar 27 at 15:15
$begingroup$
I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
$endgroup$
– Mike Earnest
Mar 28 at 16:29
$begingroup$
I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
$endgroup$
– Mike Earnest
Mar 28 at 16:29
$begingroup$
Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
$endgroup$
– Mike Earnest
Mar 28 at 16:33
$begingroup$
Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
$endgroup$
– Mike Earnest
Mar 28 at 16:33
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164173%2finverse-reverse-of-number-of-permutations-and-of-number-of-combinations-with-rep%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164173%2finverse-reverse-of-number-of-permutations-and-of-number-of-combinations-with-rep%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Question: when we are solving for $k$, is $n$ fixed?
$endgroup$
– Eevee Trainer
Mar 27 at 7:52
$begingroup$
Yes, @Eevee Trainer, when solving for k, you can assume n is fixed/known.
$endgroup$
– Brian Kennedy
Mar 27 at 15:15
$begingroup$
I think you misunderstood the algorithm you linked; it is not solving for just $n$, it solving for $k$ and $n$ simultaneously. So the code given there takes care of your first problem.
$endgroup$
– Mike Earnest
Mar 28 at 16:29
$begingroup$
Also your last problem; given the value $X$ in $X=binomn+k-1k$, apply the $C$-inverse algorithm to find $m,k$ so $X=binommk$, then let $n=m-k+1$.
$endgroup$
– Mike Earnest
Mar 28 at 16:33